Integral of sin^3(x)cos^2(x) (substitution)

[@IntegralsForYou]

🔍 𝐀𝐫𝐞 𝐲𝐨𝐮 𝐥𝐨𝐨𝐤𝐢𝐧𝐠 𝐟𝐨𝐫 𝐚 𝐩𝐚𝐫𝐭𝐢𝐜𝐮𝐥𝐚𝐫 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥? 𝐅𝐢𝐧𝐝 𝐢𝐭 𝐰𝐢𝐭𝐡 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥 𝐬𝐞𝐚𝐫𝐜𝐡𝐞𝐫: ► Integral searcher 👉integralsforyou.com/integral-searcher 🎓 𝐇𝐚𝐯𝐞 𝐲𝐨𝐮 𝐣𝐮𝐬𝐭 𝐥𝐞𝐚𝐫𝐧𝐞𝐝 𝐚𝐧 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐭𝐢𝐨𝐧 𝐦𝐞𝐭𝐡𝐨𝐝? 𝐅𝐢𝐧𝐝 𝐞𝐚𝐬𝐲, 𝐦𝐞𝐝𝐢𝐮𝐦 𝐚𝐧𝐝 𝐡𝐢𝐠𝐡 𝐥𝐞𝐯𝐞𝐥 𝐞𝐱𝐚𝐦𝐩𝐥𝐞𝐬 𝐡𝐞𝐫𝐞: ► Integration by parts 👉integralsforyou.com/integration-methods/integration-by-parts ► Integration by substitution 👉integralsforyou.com/integration-methods/integration-by-substitution ► Integration by trig substitution 👉integralsforyou.com/integration-methods/integration-by-trig-substitution ► Integration by Weierstrass substitution 👉integralsforyou.com/integration-methods/integration-by-weierstrass-substitution ► Integration by partial fraction decomposition 👉integralsforyou.com/integration-methods/integration-by-partial-fraction-decomposition 👋 𝐅𝐨𝐥𝐥𝐨𝐰 @𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥𝐬𝐟𝐨𝐫𝐲𝐨𝐮 𝐟𝐨𝐫 𝐚 𝐝𝐚𝐢𝐥𝐲 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥! 😉 📸 instagram.com/integralsforyou/

[@merylkk4hfg120]

Once again you made a problem that was confusing me look easy. Thanks!

[@IntegralsForYou]

Thanks Meryl! I am happy to see how you are enjoying the channel! ;-D

[@mrwhosetheboss3988]

❤❤❤❤4 yrs and it's still helping a lot of students..

[@IntegralsForYou]

And very proud of it! 😊

[@gdjkgdjk2523]

@@IntegralsForYou Thank you for helping me🙏

[@IntegralsForYou]

@gdjk gdjk You're welcome! 😊

[@Vishahunna55]

in the second line why did u add a negative in front of the integral? and how were u able to change sin(x0 from the first line to to -sin(x) in the second line?

[@IntegralsForYou]

Hi! I did it both actions in order to have a minus in front of sin(x) because I knew that I will do the u=cos(x) substitution which implies having du=(-sin(x))dx. If you don't want to do it, you can do u=cos(x) directly: Integral of sin^3(x)cos^2(x) dx = = Integral of sin^2(x)cos^2(x) sin(x)dx = = Integral of (1-cos^2(x))cos^2(x) sin(x)dx = Substitution: u = cos(x) du = -sin(x)dx ==> -du = sin(x)dx = Integral of (1-u^2)u^2 (-du) = = Integral of (u^2-1)u^2 du = = Integral of (u^4 - u^2) du = = u^5/5 - u^3/3 = = (1/5)cos^5(x) - (1/3)cos^3(x) + C ;-D

[@winproduction7585]

@@IntegralsForYou Thank you

[@IntegralsForYou]

@Win production My pleasure! 💪

[@me-sg8gm]

​@@IntegralsForYouThankk youu so much, i had the same doubt about This

[@dildobaggins2759]

My bone to pick with this problem is why dont we do something simpler and pick u as sin^3x but it dosent give the correct answer even though its debatebly the easier way to do the problem. Why is this?

[@IntegralsForYou]

Hi! The problem for any u-substitution is that we always have to think in expressing dx in terms of "u". It is not impossible, but is more much longer than doing u=cos(x): u = sin^3(x) ==> u^(1/3) = sin(x) 1 = sin^2(x)+cos^2(x) ==> cos^2(x) = 1-sin^2(x) = 1-u^2 du = 3sin^2(x)cos(x) dx = 3u^(2/3)sqrt(1-u^2) dx ==> du/3u^(2/3)sqrt(1-u^2) = dx Integral of sin^3(x)cos^2(x) dx = = Integral of u^(1/3)(1-u^2) du/3u^(2/3)sqrt(1-u^2) = = (1/3)Integral of u^(-1/3)(1-u^2)/u^(2/3)sqrt(1-u^2) du = =... Same idea, but shorter example: Integral of sin^3(x)cos(x) dx = = Integral of u*sqrt(1-u^2)*du/3u^(2/3)sqrt(1-u^2)) = = (1/3)Integral of u^(1/3) du = = (1/3)u^(1/3+1)/(1/3+1) = = (1/3)u^(4/3)/(4/3) = = (1/3)(3/4)u^(4/3) = = (1/4)u^(4/3) = = (1/4)(u^(1/3))^4 = = (1/4)(sin(x))^4 = = (1/4)sin^4(x) + C

[@dildobaggins2759]

@@IntegralsForYou oh yeh i got you bro cheers sin^3x is 3sin^2xcosx not 3sin^2x. That was a silly mistake of mine...

[@IntegralsForYou]

@dildo baggins No problem! 😉👍

[@karangargote3443]

Thanks 🙏🏻😊

[@IntegralsForYou]

My pleasure! ☺

[@castillomarkjayson5373]

hello what if the problem goes like this ∫cos^2(2x)sin^3(2x) dx? will it have the same answer? ty

[@IntegralsForYou]

You just have to do a little substitution before: Integral of cos^2(2x)sin^3(2x) dx = Substitution: t = 2x dt = 2 dx ==> dt/2 = dx = Integral of cos^2(t)sin^3(t) dt/2 = = (1/2)Integral of cos^2(t)sin^3(t) dt = = (1/2)*(watch video again) = = (1/2)*(-cos^3(t)/3 + cos^5(t)/5) = = (-1/6)cos^3(2x) + (1/10)cos^5(2x) + C 😉

[@tsknslhaa]

Hello, what if the x inside the parenthesis is changed into 2x, what Will be the solution? Like sin^3(x).cos^2(2x) Only one parenthesis

[@IntegralsForYou]

Of course! Here you have the solution: Integral of sin^3(x)*cos^2(2x) dx = cos(2x) = 2sin^2(x) - 1 = Integral of sin^3(x)*(2sin^2(x) - 1) dx = = Integral of (2sin^5(x) - sin^3(x)) dx = = 2*Integral of sin^5(x) dx - Integral of sin^3(x) dx = = 2* integralsforyou.com/integrals?v=w__RvsAO1BQ - integralsforyou.com/integrals?v=kBHzHDDgoIY = = 2*[ -cos(x) + (2/3)cos^3(x) - (1/5)cos^5(x) ] - [ -cos(x) - (1/3)cos^3(x) ] = = -2cos(x) + (4/3)cos^3(x) - (2/5)cos^5(x) + cos(x) + (1/3)cos^3(x) = = -cos(x) + (5/3)cos^3(x) - (2/5)cos^5(x) + C Hope I didn't do any mistake hehe

[@tsknslhaa]

@@IntegralsForYou thank u very much

[@IntegralsForYou]

@@tsknslhaa My pleasure! ☺

[@stevo0]

How can sin(x) just turn into -sin(x)?

[@IntegralsForYou]

Hi Steven Jaimes! If you are talking about the cos(x) which turned into -sin(x) (minute 1:30) it is because we derivate when we do substitution: "u = f(x)" and "du = f'(x) dx" where f'(x) is the derivative of f(x).

[@stevo0]

no not that, after the 2nd =, u have integral of( 1-cos^2(x)) (cos^2(x)) (sin(x))....then u drop down and have: negative integral of (1-cos^2(x))(cos^2(x)) (-sin(x)) dx...where does the negative before the sin come from? im guessing u multiplied the whole thing by negative? thats why theres also negative at the beginning as well? and if that is the case, wouldnt that negative at the beginning have to be distributed into everything?

[@IntegralsForYou]

As you say, the negative sign comes from multiplying by -1 inside and outside the integral, so we put a minus outside an a minus before the sin(x). I do that because I want to have -sin(x)dx and not sin(x)dx because it is the derivative of cos(x). The outsider minus is distrubuted in the last step. We do: ... = = - Integral of (u^2-u^4) du = (sign not still distributed) = - (u^3/3 - u^5/5) = (sign not still distributed) = - u^3/3 + u^5/5 = (sign distributed!) = - cos^3(x)/3 + cos^5(x)/5 + C But that doesn't mean you have to distribute it in that step. You can do it when you want, you will have exactly the same answer: Integral of sin^3(x)cos^2(x) dx = = Integral of (1-cos^2(x))cos^2(x) sin(x) dx = And from here you can do as you want: 1: = - Integral of (1-cos^2(x))cos^2(x) (-sin(x)) dx = = - Integral of (1-u^2)u^2 du = = - Integral of (u^2-u^4) du = = - (u^3/3 - u^5/5) = = - u^3/3 + u^5/5 = = - cos^3(x)/3 + cos^5(x)/5 + C 2: = Integral of -(1-cos^2(x))cos^2(x) (-sin(x)) dx = = Integral of -(1-u^2)u^2 du = = Integral of -(u^2-u^4) du = = Integral of (-u^2+u^4) du = = - u^3/3 + u^5/5 = = - cos^3(x)/3 + cos^5(x)/5 + C 3: = Integral of (-1+cos^2(x))cos^2(x) (-sin(x)) dx = = Integral of (-1+u^2)u^2 du = = Integral of (-u^2+u^4) du = = - u^3/3 + u^5/5 = = - cos^3(x)/3 + cos^5(x)/5 + C 4: = Integral of (1-cos^2(x))(-cos^2(x)) (-sin(x)) dx = = Integral of (1-u^2)(-u^2) du = = Integral of (-u^2+u^4) du = = - u^3/3 + u^5/5 = = - cos^3(x)/3 + cos^5(x)/5 + C I hope it is clear now. If not, ask again, please! ;-D

[@IntegralsForYou]

Oh, and there is another way to get the minus, if you are interested in: Integral of sin^3(x)cos^2(x) dx = = Integral of sin^2(x)cos^2(x) sin(x)dx = = Integral of (1-cos^2(x))cos^2(x)sin(x)dx = Substitution: u = cos(x) du = - sin(x)dx ==> -du = sin(x)dx = Integral of (1-u^2)(u^2)(-du) = = - Integral of (1-u^2)(u^2) du = = .....

[@vikasmalhotra7335]

Which pen are you using bro? By the way thanks for the solutions keep uploading 🙏

[@IntegralsForYou]

Hi, Vikas! It is just a black bic cristal pen ;-D Thank you for your comment! ;-D

[@annetherese]

hello,, what if the x inside the parenthesis is changed into 2x, what will be the solution? sin^3(2x)cos^2(2x)dx

[@IntegralsForYou]

Integral of sin^3(2x)cos^2(2x) dx = Substitution: t = 2x dt = 2 dx ==> dt/2 = dx = Integral of sin^3(t)cos^2(t) dt/2 = = (1/2)Integral of sin^3(t)cos^2(t) dt = = (1/2) kzbin.info/www/bejne/d6vJf519aN6Ci6s = = (1/2)( -cos^3(t)/3 + cos^5(t)/5 ) = = (-1/6)cos^3(2x) + (1/10)cos^5(2x) + C Hope it helped! ;-D

[@លូឈីទីក]

I like Math when i know how to solve it but i hate it when i don't 😂

[@IntegralsForYou]

Me too... 🤫

[@pakmai9462]

Thank you so much but why did we substitute the sin^2(x) into [1-cos(x)] instead of substituting cos^2(x) into [1-sin(x)]? And do we need to do it this way or it doesn't matter?

[@IntegralsForYou]

For the integrals sin^n(x)cos^m(x) with "n" or "m" odd integers, you have to do this substitution for the one which has the odd number. If both are odd numbers, it doesn't matter which one you choose. I think you will be interested in a playlist I did: Integrals of sin^n(x)cos^m(x) dx 👉kzbin.info/aero/PLpfQkODxXi492bpysQl1g02rPIniquUBL

[@pakmai9462]

Integrals ForYou thanks!

[@IntegralsForYou]

You're welcome! Enjoy the channel! 😀

[@hussamalanesi7432]

@@IntegralsForYou This information has helped me a lot thanks, but what if both "n" and "m" are even?

[@cnichels]

what do i need to do to remember this process

[@IntegralsForYou]

Integral of sin^n(x)cos^m(x) dx: - If n odd and m even (same for n even and odd): You can peel off one copy of that function to be your "du", and rewrite all the other copies using the basic Pythagorean identity, sin^2(x) + cos^2(x) = 1 Example: Integral of sin^3(x)cos^2(x) dx = kzbin.info/www/bejne/d6vJf519aN6Ci6s - If both n and m are equal and even: You can use the half-angle formula. sin(2x) = 2sin(x)cos(x) ==> (1/2)sin(2x) = sin(x)cos(x) ==> (sin(x)cos(x))^n = [(1/2)sin(2x)]^n = ((1/2)^n)sin^n(x) Example: Integral of sin^2(x)cos^2(x) dx = kzbin.info/www/bejne/gnO6k4ScjJlghrM These are the most common cases and of course, when the exponents are greater the integral becomes very long...

[@swattt4079]

@@IntegralsForYou thankssss

[@IntegralsForYou]

@swat tt You're welcome! 💪

[@GhanshyamSingh-ld8xe]

*Can we also do like this:* Taking sin^2 x =t Then, dwrt x, 2sin x . cosx = dt/dx Sin x . cos x dx = dt/2 So, placing these in question, Integral of t/2 . dt Then comes, ( t^2)/4 + C Comes out as (sin^4 x)/4 +C Please tell, this is correct or not as early as possible. *Please.*

[@IntegralsForYou]

Sorry for my late answer Ghanshyam Singh... You forgot a cos(x) in terms of "t" so your method doesn't work: Integral of sin^3(x)cos^2(x) dx = = Integral of sin^2(x)cos(x) sin(x)cos(x)dx = = Integral of t*cos(x)*dt/2 = ... If it is sin^3(x)cos(x) it works: Integral of sin^3(x)cos(x) dx = = Integral of sin^2(x) sin(x)cos(x)dx = = Integral of t dt/2 = = (1/2)Integral of t dt = = (1/2)(t^2)/2 = = (1/4)(t^2) = = (1/4)([sin^2(x)]^2) = = (1/4)sin^4(x) + C

[@dayzezrof2pbrawlstars171]

Like 700 This guy go me as his 700 th like after six years bro the world is cruel >:-(

[@IntegralsForYou]

Your like has been as welcomed as if it was the first one! 😊Enjoy the channel! 💪

[@soymeponguy222]

Thank you I just came here to make sure that what I did was correct and it is ☺️

[@IntegralsForYou]

Cool! It is nice to see how you use my videos! Enjoy the channel! 😀

[@Sakura-sh9du]

Thank youuuuuu

[@IntegralsForYou]

My pleasure! 😊

[@houtan4238]

At the end of the first line you forgot the "dx" but thanks it helped a lot

[@IntegralsForYou]

Oops! Sorry about that!

[@gauthamkulal3505]

In the second line why did you take u = cosx. Qsn is about sinx......

[@IntegralsForYou]

Hi! We take u=cos(x) because we have the entire expression in terms of cos(x) except for the sin(x) which is multiplying the dx. We need the sin(x) because when we do u=cos(x) then du = -sin(x)dx. In consequence, the sin(x)dx has to be substituted by the -du: Integral of sin^3(x)cos^2(x) dx = = Integral of (1-cos^2(x))cos^2(x) sin(x)dx = = - Integral of (1-cos^2(x))cos^2(x) (-sin(x))dx = Substitution: u = cos(x) du = (-sin(x))dx = - Integral of (1-u^2)u^2 du = = ... Hope it helped! ;-D

[@ahmedalhisaie7698]

Why the integral has got minus sign in the second line at 1:20 ?!

[@IntegralsForYou]

Hi! Because I wanted -sin(x)dx instead of sin(x)dx inside the integral because the derivative of cos(x) is -sin(x). You can do it in two different ways: 1. As I do in the video (multiplying by -1 inside and outside the integral): ... = Integral of (1-cos^2(x))cos^2(x)sin(x)dx = = - Integral of (1-cos^2(x))cos^2(x)(-sin(x))dx = Substitution: u=cos(x) du=-sin(x)dx = - Integral of (1-u^2)u^2 du = = ... 2. Without multiplying by -1 inside and outside the integral: ... = Integral of (1-cos^2(x))cos^2(x)sin(x)dx = Substitution: u=cos(x) du=-sin(x)dx ==> -du = sin(x)dx = Integral of (1-u^2)u^2 (-du) = = - Integral of (1-u^2)u^2 du = = ... Both methods are correct, but you can only choose one of them :-D

[@altamash198]

You make it look easy man! Thanks

[@IntegralsForYou]

:D You're welcome ;D

[@Steve-vk2kw]

Ohh man.....that was so easy!!!!! How was i not able to do this?? 😢

[@IntegralsForYou]

😜

[@Steve-vk2kw]

@@IntegralsForYou 😒

[@ashIesha]

THANK YOU !!!!!!!!!!!!!!!!!!!!!!!!!!!

[@IntegralsForYou]

My pleasure! 🙂 Enjoy the channel and the website! 🙂 integralsforyou.com/

[@zubeirjamaa]

My professor just gave this exact question on a pop quiz and I’m shocked I got it right! So happy. Only thing is I forgot the “+C” but I hope he forgives that 😅

[@IntegralsForYou]

Nooooo, you forgot the +C!! 😥😥😆 Thanks for your comment, I really appreciate it!

[@Simon-vn4jx]

thanks!

[@IntegralsForYou]

My pleasure! 😊

[@raeceljeanmaputibantilan]

The antiderivative of cosx is sinx not -sinx. Just sharing

[@IntegralsForYou]

Hi! Be careful, in 1:28 I am derivating cos(x) and not integrating it 😉

[@ТатьянаБелоусова-ь3и]

Здравствуйте, а откуда минус взялся?

[@IntegralsForYou]

Hi! Since the derivative of cos(x) is -sin(x), we need a -sin(x) multiplying the dx. If you don't like this method maybe you will like this way: ... = Integral of (1-cos^2(x))cos^2(x)sin(x)dx = Substitution: u = cos(x) du = -sin(x)dx ==> -du = sin(x)dx = Integral of (1-u^2)(u^2)(-du) = = - Integral of (1-u^2)(u^2) du = = ... Hope it helped! ;-D

[@johnsmith-hs8oi]

@@IntegralsForYou yes i understood this method better, just a suggestion but maybe you could talk through your working. Slow people like me would find it much easier to follow :)

[@tiagossoliveira]

thanks

[@IntegralsForYou]

My pleasure! ❤

[@josueisai1232]

Agradezco :3

[@IntegralsForYou]

😊

[@wexer82]

Thank you

[@IntegralsForYou]

;-D You're welcome, wexer82!

[@juliajanecek4]

oh my gosh thank you. had no idea what to do with the negative signs!!!

[@IntegralsForYou]

Hi Julia Rose Janecek! There is another way to understand these negative signs when we do substitution: Integral of sin^3(x)cos^2(x) dx = = Integral of sin^2(x)cos^2(x) sin(x)dx = = Integral of (1-cos^2(x))cos^2(x)sin(x)dx = Substitution: u = cos(x) du = - sin(x)dx ==> -du = sin(x)dx = Integral of (1-u^2)(u^2)(-du) = = - Integral of (1-u^2)(u^2) du = = ..... Thanks for your comment! ;-D

[@vihoang1499]

Many tks !!

[@IntegralsForYou]

You're welcome,Vi Hoàng! ;-D

[@majidmotamedimehr6937]

❤❤❤

[@IntegralsForYou]

❤

[@jackaitch6159]

legend

[@IntegralsForYou]

;-D

[@brainandrie2874]

Thanks, you solve my problem.

[@IntegralsForYou]

You're welcome! ;-D

[@tanujawankhede4500]

sin ^3x .cos2x

[@IntegralsForYou]

Hi Tanuja Wankhede! Here it is: cos(2x) = cos^2(x)-sin^2(x) Integral of sin^3(x)cos(2x) dx = = Integral sin^3(x)[cos^2(x)-sin^2(x)] dx = = Integral of [ sin^3(x)cos^2(x) - sin^5(x) ] dx = = Integral of sin^3(x)cos^2(x) dx - Integral of sin^5(x) dx = = kzbin.info/www/bejne/d6vJf519aN6Ci6s - kzbin.info/www/bejne/rZDCg6mpdrRkeLM dx = = -cos^3(x)/3 + cos^5(x)/5 - [ - cos(x) + 2cos^3(x)/3 - cos^5(x)/5 ] = = -cos^3(x)/3 + cos^5(x)/5 + cos(x) - 2cos^3(x)/3 + cos^5(x)/5 = = cos(x) - cos^3(x) + 2cos^5(x)/5 + C ;-D

[@stylingintro7280]

thnks

[@IntegralsForYou]

:-D

[@stunterneeraj4847]

Aur small likho